### Access Answers to NCERT Class 7 Maths Chapter 6 – The Triangle and its Properties Exercise 6.2

**1. Find the value of the unknown exterior angle x in the following diagram:**

**(i)**

**Solution:-**

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 50^{o }+ 70^{o}

= x = 120^{o}

**(ii)**

**Solution:-**

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 65^{o }+ 45^{o}

= x = 110^{o}

**(iii)**

**Solution:-**

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 30^{o }+ 40^{o}

= x = 70^{o}

**(iv)**

**Solution:-**

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 60^{o }+ 60^{o}

= x = 120^{o}

**(v)**

**Solution:-**

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 50^{o }+ 50^{o}

= x = 100^{o}

**(vi)**

**Solution:-**

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 30^{o }+ 60^{o}

= x = 90^{o}

**2. Find the value of the unknown interior angle x in the following figures:**

**(i)**

**Solution:-**

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x + 50^{o} = 115^{o}

By transposing 50^{o} from LHS to RHS it becomes – 50^{o}

= x = 115^{o} – 50^{o}

= x = 65^{o}

**(ii)**

**Solution:-**

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= 70^{o} + x = 100^{o}

By transposing 70^{o} from LHS to RHS it becomes – 70^{o}

= x = 100^{o} – 70^{o}

= x = 30^{o}

**(iii)**

**Solution:-**

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

The given triangle is a right angled triangle. So the angle opposite to the x is 90^{o}.

= x + 90^{o} = 125^{o}

By transposing 90^{o} from LHS to RHS it becomes – 90^{o}

= x = 125^{o} – 90^{o}

= x = 35^{o}

**(iv)**

**Solution:-**

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x + 60^{o} = 120^{o}

By transposing 60^{o} from LHS to RHS it becomes – 60^{o}

= x = 120^{o} – 60^{o}

= x = 60^{o}

**(v)**

**Solution:-**

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

The given triangle is a right angled triangle. So the angle opposite to the x is 90^{o}.

= x + 30^{o} = 80^{o}

By transposing 30^{o} from LHS to RHS it becomes – 30^{o}

= x = 80^{o} – 30^{o}

= x = 50^{o}

**(vi)**

**Solution:-**

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

The given triangle is a right angled triangle. So the angle opposite to the x is 90^{o}.

= x + 35^{o} = 75^{o}

By transposing 35^{o} from LHS to RHS it becomes – 35^{o}

= x = 75^{o} – 35^{o}

= x = 40^{o}

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